**1) Let G be a group and a family of subgroups. Then for any .
Proof-**

**3) The following conditions on a finite group G are equivalent.
(i) |G| is prime
(ii) and G has no proper subgroups.
(iii) for some prime p.
Proof-** Assume that the order of G is a prime, p. Then G has at least 2 elements. Pick an element a, which is not the identity and consider . By Lagrange, we know the order of a divides the order of the group. Thus the order of a is either 1 or p, but it is not 1 since a is not the identity. Thus a generates G, so G is cyclic of order p and hence isomorphic to since any finite cyclic group of a certain order is actually the same.

Assume that for some prime p. Then the order of G is p which is greater than or equal to 2, so G is nontrivial. Let . By Lagrange again, the order of H divides the order of G, so that the order of H is either 1 or p. The first case gives H to be trivial, while the second gives H to be the entire group. Thus G has no proper subgroups.

Assume that G is not trivial and G has no proper subgroups. Since G is nontrivial, there is some a in G such that a is not the identity. Since G has no proper subgroups, so that G is cyclic and of finite order, hence isomorphic to for some integer m. Suppose m is not prime then there is some natural number k such that k divides m and k is not 1 or m. Then because G is cyclic and k divides the order of G, there is exactly 1 subgroup of order k. However we know G has no proper subgroups, so this is a contradiction. Therefore m is prime.

Trivial

**4) Let and p be a prime such that . Then .
Proof-** By the division algorithm for some with . Thus taking the mod of both sides by p we get . Recall that the nonzero elements of form a group of order p-1 under multiplication mod p. Since r is in this group. Therefore the order of r, call it n, must divide p-1, the order of the group. Thus for some integer m. So we have:

but

so that

**Assume that . We prove that by subset inclusion. Let Then g can be written for some . Since as well. Note, so that we know . This tells us that there is some such that . Since H and K are subgroups of G, . Putting everything together we have that:**

6) Let H, K be subgroups of a group G. Then

Proof-

6) Let H, K be subgroups of a group G. Then

Proof-

.

Now let so that for some . Then we have that . Since H and K are subgroups we have and , so that and HK is a subgroup so that also. Thus HK = KH

Assume now that HK = KH and show . Since H and K are subgroups of G, we have that the so that which makes HK nonempty. Now we use the subgroup test:

Let Then Note that so that we can write Then we have:

. So HK is a subgroup of G.

**7) Let G be a group s.t. where p is prime and gcd(p,m) = 1. Let H, K be supgroups of G s.t. with and K not a subgroup of H. Show that
Proof- **(By contradiction) Assume that then we get . We know that the intersection of subgroups is a subgroup so so that in order for it must be the case that for some . However since K is not a subgroup of H, the intersection of H and K must contain strictly less elements than K. Thus:

so that s < d. Now we have that . By Lagrange the order of the subgroup must divide the order of G, so that which is the same as saying, for some that:

This is a contradiction because gcd(p,m) = 1. Thus HK is not a subgroup of G.

**First note two facts: and so by theorem 4.5 we can say:**

8) If H and K are subgroups of finite index of a group G such that are relatively prime, then G = HK.

Proof-

8) If H and K are subgroups of finite index of a group G such that are relatively prime, then G = HK.

Proof-

and

. This tells us that is a common multiple of both . However because we have that . Then clearly because is a common multiple it must be larger than or equal to the least common multiple so that . Lastly, by proposition 4.9 we have . Therefore so again by proposition 4.9 we can say that G = HK.

**9) If H, K, and N are subgroups of a group G such that , then
Proof- **By subset inclusion. Let Then we have . From this we can say for some . So Since H is given as a subgroup of N this means . so that by closure. But so that Thus:

since .

Now let This says that where . Since this tells us g is in K we have . Note H is a subgroup of N so that h is in N, and also g is in N. Thus their product is in N. So in conclusion we have

**10) Let H,K,N be subgroups of a group G such that . Show that H = K.
Proof-** H is already a subgroup of K so we need only show that K is inside of H. Let . Then k can be written as which says that for some we have . Then . Also we have because H is a subgroup of K. Therefore we have so that.Note the intersection of H and N is inside of H, so that . Thus . So H = K.

**11) Let G be a group of order 2n, then G contains an element of order 2. If n is odd and G is abelian, there is only one element of order 2.
Proof-** Let the order of G be 2n. Consider the set . If there does not exist an element such that then the order of this set is , which is a contradiction so there must exist at least one element of order 2.

If G is abelian and n is odd, suppose we have two elements such that . G is abelian so that is a subgroup of G. But because n is odd, we know that 4 does not divide 2n, so this subgroup cannot exist by Lagrange. Thus there can be only one element of order 2.